-(3c+4)=4(c+5)+c

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Solution for -(3c+4)=4(c+5)+c equation: 



-(3c+4)=4(c+5)+c

We move all terms to the left:

-(3c+4)-(4(c+5)+c)=0

We get rid of parentheses

-3c-(4(c+5)+c)-4=0



We calculate terms in parentheses: -(4(c+5)+c), so:

4(c+5)+c

We add all the numbers together, and all the variables

c+4(c+5)

We multiply parentheses

c+4c+20

We add all the numbers together, and all the variables

5c+20

Back to the equation:

-(5c+20)



We get rid of parentheses

-3c-5c-20-4=0

We add all the numbers together, and all the variables

-8c-24=0

We move all terms containing c to the left, all other terms to the right

-8c=24

c=24/-8

c=-3

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